= Solution
The two nonzero stationary points merge when their discriminant vanishes, giving the upper <spinodal>
$$
\boxed{T_1=T_c+\frac{b^2}{4ac},\qquad \kappa_1=\frac14}.
$$
At a nonzero stationary point, $a(T-T_c)=bP-cP^2$, so its <free energy> reduces to $F=P^3(b/6-cP/4)$. Equality with $F(0)=0$ therefore requires $P=2b/(3c)$. Substitution gives
$$
\boxed{T_2=T_c+\frac{2b^2}{9ac},\quad \kappa_2=\frac29,\quad P=0\ \text{or}\ \frac{2b}{3c}}.
$$
Both have positive curvature $2b^2/(9c)$ at coexistence. The intervening maximum is at $P=b/(3c)$. For $T_2<T<T_1$ the positive-polarization minimum is <metastable>; for $T_c<T<T_2$ the zero-polarization minimum is <metastable>.
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